What is the extreme f(x)=-x^3+3x^2+2 ?

The extreme f(x)=-x^3+3x^2+2 is Minimum(0,2),Maximum(2,6)

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extreme f(x)=-x^3+3x^2+2

extreme points

f (x) = - x^{3} + 3 x^{2} + 2

Minimum (0, 2), Maximum (2, 6)

Solution steps

f^{'} (x) = - 3 x^{2} + 6 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- x^{3} + 3 x^{2} + 2 : 2

Minimum (0, 2)

Plug

x = 2

into

- x^{3} + 3 x^{2} + 2 : 6

Maximum (2, 6)

Minimum (0, 2), Maximum (2, 6)

f (x, y) = \frac{( x ^{2} + y ^{2} - 1 )}{4}

y = 2 x^{3} - 33 x^{2} + 168 x - 8

y = x^{4} - 6 x^{2}

f (x, y) = 2 x^{2} + y^{2} - x y

f (x) = \frac{( x ^{2} + 1 )}{x}

What is the extreme f(x)=-x^3+3x^2+2 ?
The extreme f(x)=-x^3+3x^2+2 is Minimum(0,2),Maximum(2,6)